Young's modulus of steel is Y and its rigidity modulus is `eta`. A piece of steel of cross-sectional area A, is stretched into a wire of length L and area of cross-section `(A)/(4)`, In wire case

Rigidity modulus of steel is eta and its Young's modulus is Y. A piece of steel of cross sectional area A is chaged into a wire of length L and area (A)/(10) then :

Modulus of rigidity of steel is n and its Youngs's modulus is Y. A steel wire of cross sectional area A is so elongated that its area of cross section becomes A/10 . As a result

Y is the Young's modulus of the material of a wire of length L and cross-sectional area A. It is stretched through a length l. What is the force constant of the wire?

If Y is the Young's modulus of a wire of cross sectional area A, then the force required to increase its length by 0.1% will be

A force F doubles the length of wire of cross-section a The Young modulus of wire is

A force F doubles the length of wire of cross-section a The Young modulus of wire is

Young's modules of material of a wire of length ' "L" ' and cross-sectional area "A" is "Y" .If the length of the wire is doubled and cross-sectional area is halved then Young's modules will be :

The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of

The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weights added to the steel and brass wires must be in the ratio of

The Young's modulus of steel is twice that of brass. Two wires of the same length and of the same area of cross section, one of steel and another of brass are suspended from the same roof. If we want the lower ends of the wires to be at the same level, then the weight added to the steel and brass wires must be in the ratio of